Now we will turn Example 1-2 into a Living Example Problem (LEP) where we can vary parameters to learn their effect on the volume and/or exit concentration. We will use Polymath to solve the combined mole balance and rate law for the concentration profile.
We begin by rewriting the mole balance, Equation (E1-2.2), in Polymath notation form
\begin{align*} \rm{Mole Balances}\qquad\qquad\qquad\qquad&\dfrac{d(C_A)}{d(V)}=\dfrac{ra}{v0}\qquad\qquad\qquad\qquad\rm{(E1-3.1)}\\ &\dfrac{d(C_B)}{d(V)}=\dfrac{rb}{v0}\qquad\qquad\qquad\qquad\rm{(E1-3.2)} \end{align*}
\begin{align*} \rm{Rate Law}\qquad\qquad\qquad\qquad\qquad&ra=-k*C_A\qquad\qquad\qquad\qquad\rm{(E1-3.3)}\\ &rb=-ra\qquad\qquad\qquad\qquad\qquad\rm{(E1-3.4)} \end{align*}
The parameter values are $k = 0.23 \rm{min}^{-1}$, $v_0=10\rm{dm}^3/s$ and $C_{A0}=10mole/dm^3$
The initial and final values for the integration wrt the volume V are V=0 and V=100 $dm^3$.